Limit of a monotone sequence of measurable functions is measurable. I'm currently studying measure theory as a non mathematician out of self interest but I've come across a typical proof style that I find rather hard. Generally it's showing that a typical set can be written as a countable union of measurable sets. An example I'm having trouble with is below.
Theorem
Let $(X, \mathcal{A})$ be a measurable space and let $(f_n:n \in \mathbb{N})$ be a monotone sequence of of extended real valued $\mathcal{A}$- measurable functionson a set $D \in \mathcal{A}$. Then $\lim_{n \rightarrow \infty}f_n$ exists on $D$ and is $\mathcal{A}$- measurable on $D$.
Proof
For the increasing case. Usual idea in that we would like to write the set $\{D: \lim_{n \rightarrow \infty} f_n > \alpha\}$ as a union of measurable sets. The proof states that
\begin{equation}
\{D: \lim_{n \rightarrow \infty} f_n > \alpha\} = \bigcup_{n \in \mathbb{N}} \{D: f_n > \alpha\} \in \mathcal{A}
\end{equation}
However, I'm not particularly confident in my ability to show the equality above. The book gives the hint that $\lim_{n \rightarrow \infty} f_n > \alpha $ iff $f_n > \alpha$ for some $n \in \mathbb{N}$.
I've tried to prove this as follows:
Write $f=\lim_{n \rightarrow \infty} f_n$. Assume that $f \leq \alpha$. Since the pointwise limit $f_n(x)$ exists for each $x \in D$ by the monotone convergence theorem. Therefore, for all $n > K\in \mathbb{N}$ we have that
\begin{equation}
|f-f_n|< \epsilon 
\end{equation}
Or
\begin{equation}
-\epsilon + f_n < f < f_n + \epsilon 
\end{equation}
In particular, $f > f_n - \epsilon$ so $f \geq f_n > \alpha$ which is a contradiction. Hence $f_n > \alpha$ for some $n \in \mathbb{N}$ $\implies$ $\lim_{n \rightarrow \infty} f_n > \alpha$. I'm not so sure about the other case.
As for showing the sets are equal I'm quite happy that $x \in \bigcup_{n \in \mathbb{N}} \{D: f_n > \alpha\}$ implies that $x \in \{D: f_n > \alpha\}$ for some $n$ which from the hint above implies that $x \in \{D: \lim_{n \rightarrow \infty} f_n > \alpha\}$ hence $ \bigcup_{n \in \mathbb{N}} \{D: f_n > \alpha\} \subset \{D: \lim_{n \rightarrow \infty} f_n > \alpha\}$.
If you could provide any hints or advice I would be very grateful.
 A: It is a standard result in calculus that weak inequalities are preserved in a limit. Suppose $x\in\{D: f>\alpha\}$. If we had $x\notin\cup_{n=1}^\infty \{D: f_n>\alpha\}$ then it would mean that we have $f_n(x)\leq\alpha$ for all $n\in\mathbb{N}$. But then passing to the limit this would imply $f(x)\leq\alpha$, a contradiction.
Other direction: suppose $x\in\cup_{n=1}^\infty \{D: f_n>\alpha\}$. Then there exists $n_0\in\mathbb{N}$ such that $f_{n_0}(x)>\alpha$. By monotonicity this implies we have $f_n(x)\geq f_{n_0}(x)>\alpha$ for all $n\geq n_0$. Again, since weak inequalities are preserved in a limit we conclude that $f(x)=\lim_{n\to\infty} f_n(x)\geq f_{n_0}(x)>\alpha$. So $x\in\{D: f>\alpha\}$.
A: As pointed out in the comments, it is not the best idea to work with the $\epsilon$-version of limit. It is easier to work with the monotone convergence directly. Suppose that $f_n \uparrow f$ (otherwise look at $g_n := -f_n$).
Fix $x\in X$.
It follows directly that $f(x) \leq a$ implies $f_n(x) \leq f(x) \leq a$ for every $n$. In particular, we get that $f_n(x) \leq a$ for some $n$. That is the first inclusion you already showed.
For the second, suppose that $f_n(x) \leq a$ for every $n$. Then by monotone convergence properties, one directly gets $f(x) \leq a$.
